How do you write a quadratic equation that passes through (0,9) and vertex is (-2,5)?

Mar 17, 2016

$y = {\left(x + 2\right)}^{2} + 5$
The other form of this is: $\text{ } y = {x}^{2} + 4 x + 9$

Explanation:

Standard form is:$\text{ } y = a {x}^{2} + b x + c$

Vertex form is: $y = a {\left(x + \textcolor{red}{d}\right)}^{2} + k$.................(1)

We know it passes through two points:

${P}_{1} \to \left({x}_{1} , {y}_{1}\right) \to \left(0 , 9\right)$

$\textcolor{b r o w n}{{P}_{2} \to \left({x}_{2} , {y}_{2}\right) \to \left(\textcolor{b l u e}{- 2} , \textcolor{m a \ge n t a}{5}\right) \to V e r t e x}$

Known that ${x}_{\text{vertex}} = \left(- 1\right) \times \textcolor{red}{d} = \textcolor{b l u e}{- 2}$ (in this case)

$\implies d = + 2$
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Now we have:

$y = a {\left(x + \textcolor{red}{2}\right)}^{2} + k$.................(2)

It is also known that ${y}_{\text{vertex}} = \textcolor{m a \ge n t a}{5}$ (in this case)

Thus equation (2) becomes

$y = a {\left(x + \textcolor{red}{2}\right)}^{2} + \textcolor{m a \ge n t a}{5}$.................(2)
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Let us assume that $a = 1$ if it is not then the plot will not pass through ${P}_{2}$. As the plot does pass through this point then the completed equation is $y = {\left(x + 2\right)}^{2} + 5$

The equation can also be derived using the standard form of
$y = a {x}^{2} + b x + c$. Doing so would involve a lot more work.